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# 1110 - MATH 153 Selected Solutions for 11.10 Exercise 10...

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MATH 153 Selected Solutions for § 11.10 Exercise 10 : Find the Maclaurin series for f ( x ) using the definition of a Maclaurin series. Also find the associated radius of convergence. f ( x ) = xe x Solution : We find a few derivatives to find the pattern in f ( n ) (0): f ( x ) = xe x f (0) = 0 f 0 ( x ) = xe x + e x f (1) (0) = 1 f 00 ( x ) = xe x + 2 e x f (2) (0) = 2 f 000 ( x ) = xe x + 3 e x f (3) (0) = 3 It seems clear that f ( n ) ( x ) = xe x + ne x , so that f ( n ) (0) = n . Thus, xe x = X n =0 f ( n ) (0) n ! x n = X n =0 n n ! x n = X n =0 1 ( n - 1)! x n . To find the radius, we apply the Ratio Test: lim n →∞ x n +1 n ! x n ( n - 1)! = lim n →∞ | x | n +1 n ! · ( n - 1)! | x | n = lim n →∞ | x | n = 0 < 1 . Since the limit is always less than one regardless of x , the radius R = . Exercise 20 : Find the Taylor series for f ( x ) centered at the given value of a . f ( x ) = x - 2 , a = 1 Solution : This is the same as the last problem, except that we plug in x = a instead of x = 0. We find a few derivatives to find the pattern in f ( n ) ( a ): f ( x ) = x - 2 f (1) = 1 f 0 ( x ) = - 2 x - 3 f (1) (1) = - 2 f 00 ( x ) = ( - 3)( - 2) x - 4 = 3! x - 4 f (2) (1) = 6 = 3! f 000 ( x ) = ( - 4)( - 3)( - 2) x - 5 f (3) (1) = - 24 = - 4! So f ( n ) ( x ) = ( - 1) n ( n + 1)! x - ( n +2) , and f ( n ) (1) = ( - 1) n ( n + 1)!. Thus, x - 2 = X n =0 f ( n ) (1) n ! ( x - 1) n = X n =0 ( - 1) n ( n + 1)! n ! ( x - 1) n = X n =0 ( - 1) n ( n +1)(

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